山东省济南市2020-2021学年高二下学期期末考试物理试题答案

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高二物理参考答案一、单项选择题:本题共8个小题,每小题3分,共24分。1.B2.A3.C4.D5.B6.C7.D8.A二、多项选择题:本题包括4小题,每小题4分,共16分。9.BD10.BCD11.BC12.AD三、非选择题:本题共6个小题,满分60分。13.(1

)ABD(2分,漏选得1分,错选不得分)(2)B(2分)(3)1.16×10-2(1.14×10-2~1.18×10-2之间都对)(2分);6×10-10(2分)14.(1)BC(2分,漏选得1分,错选不得分)(2)①tn(2分)③2224πtTLgn=+

(2分)④24πk(2分)15.解:(1)A质点开始振动的方向沿y轴负方向,故第二次出现波峰经历的时间t1=314T解得T=0.4s·······························(2分)由图可得λ=4m∴10λvT==m/s················(2分)(2)B点第一

次出现波谷的时刻为209xtT.v=+=s(2分)B走过的路程为为s=5A=50cm····················(2分)16.(1)位移最大时,加速度最大,m所受静摩擦最大。对整体列牛顿第二定律方程有:()kAMma=+··················

·············(1分)∴()kAaMm=+对m有:f=ma··································(1分)µmg≥f····································(1分

)解得:()kAμMmg+····················(1分)(2)对系统,由能量守恒可得:()2221112222AkAkMmv=++···················

··(2分)解得:32AkvMm=+·····································(2分)17.解:(1)由题意可知,出射角r=45o····················

(1分)sinsinrni=·····································(2分)解得:32sin10i=······························(1分)(2)临界角1sinCn=···················

··················(2分)解得:C=37o要使光线在AB、AC界面均发生全反射,需满足37oθ90o-θ≥37o可得:37o≤θ≤53o过P点做AB垂线交于Q,由几何关系得:PQ=aQM=a3tan374oa=

································(3分)在AB边上的入射角β=53o故在ΔPQN中,由几何关系可得:QN=a4tan533oa=··············(2分)综上所述,符合要求的区域为4373412MNaaa=−=·····

····················(2分)18.(1)取上部气缸中的气体作为研究对象,由波意尔定律得:01222LpLSpS=······························①(2分)解得:p1=2p0···························

···(1分)根据120010hppp=+可得h1=10m(1分)(2)设B活塞相对气缸移动的距离为x1,对两部分气体分别由波意尔定律得:0212210LpLSpSxS=+························②(2分)()0214pLSpLx

S=−·······························③(1分)联立②③可解得:205pp=································(1分)∴h2=40m···

·················································(1分)(3)设A活塞恰好移动至上部气缸底端时水深为H,此时B活塞相对气缸移动的距离为x2,则对两部分气体分别由波意尔定律得:0322pLSpxS=···

······················④(1分)()0324pLSpLxS=−·····················⑤(1分)联立④⑤可解得:23Lx=306pp=(2分)∴H=50m································

···············(1分)此后随着水深的增加,A活塞和B活塞的位置不再便化故水深60m处,稳定后活塞B相对气缸N移动的距离23Ldx==(1分)

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